1959 AMC 12 Problem 32

Question

The length

l

of a tangent, drawn from a point

A

to a circle, is

\dfrac43

of the radius

r.

The (shortest) distance from

A

to the circle is:

\dfrac12r

r

\dfrac12l

\dfrac23l

a value between

r

and

l

Join

A

to the center and to the point of tangency

Use the right triangle with legs

r

and

l=\frac{4r}{3}

If

O

is the center and

T

the point of tangency, then

OT\perp AT.

Hence

\begin{aligned} AO &=\sqrt{r^2+l^2}\\ &=\sqrt{r^2+\frac{16r^2}{9}}\\ &=\frac{5r}{3}. \end{aligned}

The shortest distance from

A

to the circle is

AO-r=\frac{2r}{3}.

Since

l=\frac{4r}{3},

this distance equals

\frac{l}{2}.

Therefore, the correct answer is C.

Problem 32 in other years